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Let $T^2=S^1 \times S^1$ be a torus, where $S^1 \subset \mathbb{R}^2$ is the unit circle. Is $T^2$ connected? If so, how does one show this?

Philippe Knecht
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    Tell us what you tried or don't understand. – paul blart math cop Mar 02 '21 at 12:21
  • Looks connected to me. – Adam Rubinson Mar 02 '21 at 12:23
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    I was able to show the lemma ($X$ connected, $f:X\to Y$ continuous and surjective $\implies Y$ connected) as well as the lemma ($X$ connected $\iff \emptyset$ and $X$ are the only clopen sets in $X$). Intuitively suspecting that $T^2$ is connected (consider a store-bought donut for verification), I'm wondering whether there is a surjective continuous map $f:\mathbb{R}^n \to T^2$ or whether I need to use a different strategy. – Philippe Knecht Mar 02 '21 at 12:24
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    Product of connected spaces is connected; or use path-connectedness. – Chrystomath Mar 02 '21 at 12:32

2 Answers2

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I found an answer that somewhat follows the results we have seen in my topology course so far. For this we use a couple of definitions:

Definitions.

A topological space $X$ is called connected, if there does not exist any decomposition $X = U \cup V$ into non-empty, disjoint subsets $U,V \subseteq X$.

A topological space $X$ is called path-connected, if $\forall x,y \in X$ there exists a continuous map $\gamma:[a,b] \to X$ with $[a,b] \subseteq \mathbb{R}$ and $\gamma(a)=x,\,\gamma(b)=y$.

Proof. We make use of some results that can be proven independently. First, we notice that

For $n \geq 1$ we have that the $n$-sphere $S^n = \{x \in \mathbb{R}^{n+1} \mid \|x\|_2 = 1\}$ is path-connected.

Now, because of the fact

Let $X$ and $Y$ be topological spaces. $X$ and $Y$ are path-connected $\iff X \times Y$ is path-connected.

we can conclude that $S^1 \times S^1$ is path-connected. Finally,

Path-connected topological spaces are connected.

Hence, $T^2 = S^1 \times S^1$ is connected. QED.

Philippe Knecht
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One way: the torus is the continuous image of the connected space the unit square $S$ under certain identifications $\sim$. The projection $p:S\to S/\sim \cong T^2$ is a quotient map and hence continuous. (The identifications are of the opposite sides of the square.)